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This page sets out the method for conditional probability in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Conditional probability topic page.
When items are taken one after another and not put back.
When a question says 'given that' with a Venn diagram or table, only look inside the given group.
When a counter is not replaced, the total number of counters goes down by 1. The count of the colour just drawn goes down by 1 as well.
From 5 red and 3 blue (total 8), if a red is drawn first, 4 red remain out of 7, so P(second red | first red) = 4/7.
Multiply the probabilities along the branch: P(both red) = P(first red) × P(second red | first red). For 5 red and 3 blue: 5/8 × 4/7 = 20/56 = 5/14.
If the first was red, the blue count is unchanged. For 5 red and 3 blue: P(second blue | first red) = 3/7.
Give probabilities as simplified fractions. A rounded decimal such as 0.333 loses accuracy when the exact answer is 1/3.
Before any counter is removed, the probability is simply the count of that colour divided by the original total.
The formula looks like a division because you are rescaling. P(A and B) is the share of the whole group where both facts are true. P(B) is the share where the condition is true. Dividing the first by the second asks: of the times B happens, how often does A happen too? If P(A and B) = 6/30 and P(B) = 15/30, then P(A mid B) = 6/15 = 2/5. The 30 cancels. You are left with the reduced group.
Do not multiply here. Multiplying would answer a different question, and adding joins alternatives. The bar in P(A mid B) means given, so divide by P(B). If the condition has probability 0, there is no reduced group to look at.
Method: 1. Name the condition, the event after given. 2. Find the probability of both events together. 3. Divide that probability by the probability of the condition. 4. Simplify the resulting fraction.
Without replacement, the second probability is conditional on the first colour. Start with 9 red counters and 6 yellow counters, so the bag holds 15. The probability the first is red is 9/15. Given that it was red, 8 red counters remain and 14 counters remain. The second probability is 8/14. Both red is the product 9/15 × 8/14 = 12/35 after simplifying. You multiply because both draws have to be red. The second factor has already used the condition.
Leaving the second factor as 9/15 treats the draw as replacement. The counts would be restored, and the events would be independent. The question said the counter stays out, so the information from the first draw has to change the bag. Write the new numerator and the new denominator before you multiply. Then cancel. An unsimplified product such as 72/210 is the right value, but lowest terms is the form to finish in.
Method: 1. Write the first probability from the full bag. 2. Adjust the colour count for the counter that left. 3. Adjust the total by 1. 4. Multiply the first probability by that conditional probability.
Sometimes the question stops at the second draw. It tells you the first colour and asks only for the next probability. You do not multiply by the first branch, because that branch has already happened. It is certain under the condition. From 8 red and 5 blue, given that the first counter is red, the probability the second is red is 7/12. One red has left, so 7 red remain. The total was 13 and is now 12. The first fraction 8/13 is not part of the answer.
If the required colour is not the colour that left, change only the total. A bag of 11 counters has 4 green and 7 white. Given that the first counter is white, the green count is still 4, and the total is 10. The probability the second is green is 4/10 = 2/5. Reducing the green count as well would invent a green counter that never left. Read which colour was removed before you edit a numerator.
Method: 1. Note the colour that left the bag. 2. Subtract 1 from that colour only. 3. Subtract 1 from the total. 4. Put the colour you still want over the reduced total, and leave the first draw out.
A table gives the same shrinking without a tree. In a class of 30 pupils, 15 walk, and 6 of those walkers bring a packed lunch. Pick a pupil who walks. The 15 who do not walk are excluded by the condition. The probability of a packed lunch, given that the pupil walks, is 6/15 = 2/5. Using 6/30 answers a different question: the probability that a random pupil from the whole class both walks and brings lunch. The word given throws the other 15 pupils out.
The formula gives the same result. P(lunch and walk) = 6/30 and P(walk) = 15/30, so the quotient is 6/15 = 2/5. The class size cancels. On a frequency tree, stand on the branch you were given and read the next split as a fraction of that branch.
Method: 1. Find the row or the branch named by the condition. 2. Use that row total as the denominator. 3. Use the matching cell as the numerator. 4. Simplify, and do not divide by the grand total.
A conditional probability question always contains a clue such as "given that", "of those who", or "a boy is chosen, what is the probability that". The clue tells you the group you are now picking from. Every other member of the original group is out of play, so the denominator is the size of that smaller group, and the numerator is the part of it that matches the second event.
In symbols, P(A mid B) is read "the probability of A given B". It equals (P(A cap B))/(P(B)). With counts, this is the number in both events divided by the number in event B. If a question gives probabilities instead of counts, use the same fraction with probabilities, and you can scale to whole numbers if that makes the arithmetic friendlier.
Method: 1. Underline the condition and say which group you are choosing from. 2. Count the people or items in that group, and use this as the denominator. 3. Count how many of that group also match the second event, and use this as the numerator. 4. Write the fraction and simplify it.
Tip: If the denominator in your answer is the same as the whole total in the question, ask yourself whether you have used the condition.
Exam questions often hide the information in a two-way table or a Venn diagram. For a table, the condition picks out one row or one column, and the probability uses only that row or column. For a Venn diagram, the condition picks one circle. Everything outside the circle is ignored, so the denominator is the whole circle, including the overlap, and the numerator is the overlap.
Totals in the margins of a table are a useful check. If a row total is missing, find it first, then work out the probability. A common set-up is "a girl is chosen at random, what is the probability she chose the museum". The words "a girl is chosen" fix the denominator as the number of girls, not the number of students.
Method: 1. Fill in any missing totals in the table or any missing region in the Venn diagram. 2. Pick out the row, column or circle that the condition describes. 3. Use the total of that row, column or circle as the denominator. 4. Use the matching count inside it as the numerator, then simplify.
When items are taken without replacement, the second probability depends on what came first. A phrase such as "at least one red" is easier to handle through its opposite: P(at least one red) = 1 - P(no reds). The opposite is a single branch, so there is one product instead of three, and it saves time and errors.
To work out something like "given that at least one is red, the probability that both are red", find the top and the bottom separately. The top is P(both red). The bottom is P(at least one red). Divide the top by the bottom. Because both probabilities are fractions with the same denominator, the working is often short once you write them over a common denominator.
Method: 1. Write the probabilities of each branch as fractions, reducing the total by one for the second draw. 2. Find the probability of the event that is given, using 1 - P(none) if that is quicker. 3. Find the probability that both the event and the condition happen. 4. Divide the second result by the first, then simplify.
Tip: Keep the fractions unsimplified until the very end. Common denominators make the division easy.
Some of the hardest questions run time backwards: "Given that the second counter is red, what is the probability that the first counter was also red?" The second counter being red can happen two ways, first red then red, or first another colour then red. Work out both branches, add them, and use that sum as the bottom of the fraction.
The same method works when a first stage chooses a bag or a coin side, and then an item is taken from the chosen bag. List every route that leads to the given outcome, add their probabilities for the denominator, and then take the probability of the route you are asked about as the numerator. Writing a small tree, even roughly, makes the routes clear and stops you missing one.
Method: 1. List every route that ends in the given outcome. 2. Multiply along each route and add the products to get the denominator. 3. Pick out the route the question asks about as the numerator. 4. Divide numerator by denominator and simplify.
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